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I assume [tex]C[/tex] is oriented counterclockwise. If [tex]C[/tex] is the closed rectangle, then by Green's theorem we have
[tex]\displaystyle \int_C F\cdot dr = \int_0^3 \int_0^9 \frac{\partial (x-y)}{\partial x} - \frac{\partial (x^2+y^2)}{\partial y} \, dy \, dx \\\\ ~~~~~~~~~~~~ = \int_0^3 \int_0^9 (1 - 2y) \, dy \, dx = 3 \int_0^9 (1-2y)\, dy = \boxed{-216}[/tex]
It seems unlikely, but if you actually are omitting the integral along the line segment joining (0, 9) and (0, 0), let [tex]x=0[/tex] and [tex]y=9-t[/tex] with [tex]0\le t\le9[/tex]. Then the integral along this line segment would be
[tex]\displaystyle \int_{(0,9)\to(0,0)} F\cdot dr = \int_0^9 \bigg((0^2 + (9-t)^2) i + (0 - (9-t)) j\bigg) \cdot (0i - j) \, dt \\\\ ~~~~~~~~ = \int_0^9 (9-t) \, dt = \frac{81}2[/tex]
so that the overall integral would instead have -216 - 81/2 = -513/2.